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JEE Main levelSingle correct optionDifficulty 2 of 3mt200

On an Ellingham diagram the line for $2 C + O X_{2} 2 CO$ crosses the line of a metal oxide $MO$ at a temperature $T_{0}$ . For $T > T_{0}$

Athe reaction has $Δ G^{\circ} = 0$ at all higher temperatures
BM reduces $CO$ to carbon
Ccarbon cannot reduce $MO$ as the $MO$ line is lower
Dcarbon can reduce $MO$ to M because $Δ G^{\circ}$ of $MO + C M + CO$ is negative

Answer: D

The carbon line has a negative slope and goes below the oxide line (positive slope) after

T_{0}

.

Δ G^{\circ}

for

MO + C M + CO

is half the difference of the two lines and is negative above

T_{0}

.

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